Question

write the equation of the tangent line to the curve y=x^3+3x-4 at (2,10).
The equation of the tangent line is ?

Answer 1

y = f(x)

Equation for a tangent line at a point (a, f(a))

y - f(a) = f '(a)*(x - a)

Answer 2

The derivative, f '(a) is the slope of the tangent line at the point x=a

Answer 3

Calculate the derivative, then plug into that x=2, f ' (2) = ?

y - 10 = f '(2)*(x - 2)

just find the derivative evaluated at 2 and put it in the equation above for f '(2)

Answer 4

would the equation be 3x^2-3?

Answer 5

or would I plug 2 into that equation?

Answer 6

for y=f(x)
The derivative is f '(x) = 3x^2 + 3

Evaluate that at x=2, f '(2) will be the slope for your tangent line

Answer 7

If you have a slope, and a point, recall the point slope formula from algebra...
point (a , f(a))
slope = f '(a)

y - f(a) = f '(a)*(x - a)

Answer 8

I have to run... the derivative f '(x) = 3x^2 + 3

slope of tangent line at point x=2, y=10 is f '(2) = 15

Tangent line y - f(a) = f '(a)*(x - a)

y - 10 = 15*(x - 2)

rearrange if you want a different form... here is the graph

Answer 9

Ill be back on later probably..

Answer 10

so the equation becomes y=15x-20 once you simplify?

Answer 11

\[f(x)=x^3+3x-4\]\[f'(x)=3x^2+3\]

The slope of the tangent line at point (2,10) is\[f'(2)=15\]
\[y=mx+b\]And we know m = 15 for the point (2, 10). Solve for b.

y = 15x + b
10 = 15(2) + b
b = -20

The equation of the tangent line is\[y=15x-20\]